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\title{Chapter 03: Rings of Differential Operators}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
In this chapter we show that the Weyl algebras are members of the family of rings of differential operators. 

These rings come up in many areas of mathematics: representation theory of Lie algebras, singularity theory and differential equations are some of them.

\end{frame}

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% Section 1
\section{Definitions}
\begin{frame}[allowframebreaks]{A. }

\vspace{-0.5cm}

Let $R$ be a commutative $K$-algebra. 

The ring of differential operators of $R$ is defined, inductively, as a subring of $\text{End}_K(R)$. 

As in the case of the Weyl algebra, we will identify an element $a \in R$ with the operator of $\text{End}_K(R)$ defined by the rule $r \mapsto ar$, for every $r \in R$.

We now define, inductively, the order of an operator. 

An operator $P \in \text{End}_K(R)$ has {\color{red}order zero} if $[a,P] = 0$, for every $a \in R$. 

Suppose we have defined operators of order $< n$. 

An operator $P \in \text{End}_K(R)$ has {\color{red}order $n$} if it does not have order less than $n$ and $[a,P]$ has order less than $n$ for every $a \in R$. 

Let $D^n(R)$ denote the set of all operators of $\text{End}_K(R)$ of order $\leq n$. 

It is easy to check, from the definitions, that $D^n(R)$ is a $K$-vector space.

We may characterize the operators of order $\leq 1$ in terms of well-known concepts. 

A {\color{red}derivation} of the $K$-algebra $R$ is a linear operator $D$ of $R$ which satisfies Leibniz's rule: $D(ab) = aD(b) + bD(a)$ for every $a,b \in R$. 

Let $\text{Der}_K(R)$ denote the $K$-vector space of all derivations of $R$. 

Of course $\text{Der}_K(R) \subseteq \text{End}_K(R)$. 

If $D \in \text{Der}_K(R)$ and $a \in R$, we define a new derivation $aD$ by $(aD)(b) = aD(b)$ for every $b \in R$. 

The vector space $\text{Der}_K(R)$ is a left $R$-module for this action. 

Derivations are very important in commutative algebra and algebraic geometry as shown in [Matsumura 86] and [Hartshorne 77].

\textbf{Lemma 1.1.} The operators of order $\leq 1$ correspond to the elements of $\text{Der}_K(R) + R$. 

The elements of order zero are the elements of $R$.

Proof: Let $Q \in D^1(R)$ and put $P = Q - Q(1)$. 

Note that $P(1) = 0$ and that $P$ has order $\leq 1$. 

Hence $[P,a]$ has order zero for every $a \in R$. 

Thus for every $b \in R$, we have that $[[P,a],b]=0$. 

Writing the commutators explicitly, one obtains the equality
\begin{equation}
(Pa)b - (aP)b - b(Pa) + b(aP) = 0.
\end{equation}

Applying this operator to $1 \in R$, we end up with $P(ab) = aP(b) + bP(a) - baP(1)$. 

Since $P(1)=0$, it follows that $P$ is a derivation of $R$. 

But $Q = P + Q(1) \in \text{Der}_K(R) + R$, as required. 

An easy calculation shows that if $Q$ has order zero then $Q \in R$.

The {\color{red}ring of differential operators} $D(R)$ of the $K$-algebra $R$ is the set of all operators of $\text{End}_K(R)$ of finite order, with the operations of sum and composition of operators. 

In other words, $D(R)$ is the union of the $K$-vector spaces $D^n(R)$ for $n=0,1,2,\ldots$. 

For this definition to make sense, we must show that the sum and product of two operators of finite order has finite order. 

This is clear for the sum, but requires a proof for the multiplication.

\textbf{Proposition 1.2.} Let $P \in D^n(R)$ and $Q \in D^m(R)$, then $P \cdot Q \in D^{n+m}(R)$.

Proof: The proof is by induction on $m+n$. 

If $m+n=0$ the result is obvious. 

Suppose the result true whenever $m+n < k$. 

If $m+n=k$ and $a \in R$, we have that
\begin{equation}
[PQ,a] = P[Q,a] + [P,a]Q.
\end{equation}

The definition of order implies that $[Q,a] \in D^{m-1}(R)$ and $[P,a] \in D^{n-1}(R)$. 

Thus, by the induction hypothesis $P[Q,a], [P,a]Q \in D^{n+m-1}$. 

Hence $[PQ,a]$ belongs to $D^{n+m-1}$, as required.

We end this section with the explicit calculation of $\text{Der}_K(K[x_1, \ldots, x_n])$, which will be needed in the next section.

\textbf{Proposition 1.3.} Every derivation of $K[X] = K[x_1, \ldots, x_n]$ is of the form $\sum_{1}^{n} f_i \partial_i$, for some $f_1, \ldots, f_n \in K[X]$.

Proof: Let $D \in \text{Der}_K(K[X])$. 

Then $D(x_i^k) = kx_i^{k-1}D(x_i)$, for $i=1, \ldots, n$. 

Hence
\begin{equation}
(D - \sum_{1}^{n} D(x_i)\partial_i)(x_1^{a_1} \cdots x_n^{a_n}) = 0.
\end{equation}

Since these monomials form a basis for $K[X]$ we have that $D = \sum_{1}^{n} D(x_i)\partial_i$. 

Thus $D = \sum_{1}^{n} f_i\partial_i$ with $f_i = D(x_i)$.

\end{frame}

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% Section 2
\section{The Weyl Algebra}
\begin{frame}[allowframebreaks]{B. }

\vspace{-0.5cm}

Our aim now is to show that the Weyl algebra is the ring of differential operators of the algebra of polynomials. 

For the proof of this result we need two lemmas.

\textbf{Lemma 2.1.} Let $P \in D(K[X])$. 

If $[P,x_i] = 0$ for every $i=1,\ldots,n$, then $P \in K[X]$.

Proof: We will show that under this hypothesis, $[P,f]=0$ for every $f \in K[X]$. 

Then the result follows from Lemma 1.1. 

Since the commutator is additive, it is enough to prove that $[P,f]=0$ when $f$ is a monomial in $K[X]$. 

Let $f=x^{\alpha}$, for some $\alpha \in \mathbb{N}^n$ and assume that $\alpha_i \neq 0$. 

Then
\begin{equation}
[P,x^{\alpha}] = [P,x_i]x^{\alpha-\epsilon_i} + x_i[P,x^{\alpha-\epsilon_i}].
\end{equation}

By induction on the degree of the monomials, $[P,x_i]=[P,x^{\alpha-\epsilon_i}]=0$. 

Thus $[P,x^{\alpha}]=0$, as required.

The next lemma is formally equivalent to the fact that every polynomial vector field $F=(F_1,\ldots,F_n)$ in $\mathbb{R}^n$ which satisfies $\partial F_j/\partial x_i=\partial F_i/\partial x_j$ for all $1 \leq i,j \leq n$, has a {\color{red}potential}. 

Define $C_r$ to be the set of operators in $A_n$ which can be written in the form $\sum_{\alpha} f_{\alpha}\partial^{\alpha}$ with $|\alpha| \leq r$. 

A simple calculation shows that
\begin{equation}
C_r = C_{r+1} \cap D^r(K[X]).
\end{equation}

By Proposition 1.3, we have that $C_1 = \text{Der}_K(K[X]) + K[X]$ and that $C_0 = K[X]$. 

We will use the convention that if $k < n$ then $\mathbb{N}^k$ is embedded in $\mathbb{N}^n$ as the set of $n$-tuples whose last $n-k$ components are zero.

\textbf{Lemma 2.2.} Let $P_1,\ldots,P_n \in C_{r-1}$ and assume that $[P_i,x_j]=[P_j,x_i]$ whenever $1 \leq i,j \leq n$. 

Then there exists $Q \in C_r$ such that $P_i=[Q,x_i]$, for $i=1,\ldots,n$.

Proof: Suppose, by induction, that we have determined $Q' \in C_r$ such that $[Q',x_i]=P_i$ for $k+1 \leq i \leq n$; thus $[[Q',x_i],x_k]=[P_k,x_i]$. 

Write $G=[Q',x_k]-P_k$. 

Then $[G,x_i]=0$, for $k+1 \leq i \leq n$.

It follows from the identity
\begin{equation}
[\partial^{\beta}, x_n] = \beta_n \partial^{\beta-\epsilon_n}
\end{equation}

that if $[\partial^{\beta}, x_n]=0$ then $\beta_n=0$. 

Thus $[G,x_n]=0$ implies that $G$ can be written as a linear combination of monomials of the form $x^{\alpha}\partial^{\beta}$ with $\beta \in \mathbb{N}^{n-1}$. 

Since $[G,x_i]=0$ for $k+1 \leq i \leq n$, we may apply this result several times and conclude that
\begin{equation}
G = \sum_{\alpha \in \mathbb{N}^k} f_{\alpha} \partial^{\alpha}
\end{equation}
where $f_{\alpha} \in K[X]$. 

Now write
\begin{equation}
Q'' = \sum_{\alpha \in \mathbb{N}^k} (\alpha_k + 1)\pow{-1} f_{\alpha} \partial^{\alpha+\epsilon_k}.
\end{equation}

However, $Q' \in C_r \subseteq D^r(K[X])$ implies that
\begin{equation}
[Q',x_k] \in C_r \cap D^{r-1}(K[X]) = C_{r-1}.
\end{equation}

Since $P_k$ also belongs to $C_{r-1}$, then so does $G$. 

Hence $Q'' \in C_r$. 

On the other hand, $[Q'',x_i]=0$ for $k+1 \leq i \leq n$ by construction. 

Thus $[Q'-Q'',x_i]=P_i$. 

But $[Q'',x_k]=G$, and so
\begin{equation}
[Q'-Q'',x_k] = [Q',x_k] - G = P_k
\end{equation}

Hence, $[Q'-Q'',x_i]=P_i$, for $k \leq i \leq n$; and the induction is complete.

\textbf{Theorem 2.3.} The ring of differential operators of $K[X]$ is $A_n(K)$. 

Besides this, $D^k(K[X])=C_k$.

Proof: It is enough to prove that $D^k(K[X]) \subseteq C_k$. 

Let $P \in D(K[X])$. 

If $P \in D^1(K[X])$ then by Lemma 1.1, $P \in \text{Der}_K(K[X]) + K[X]$. 

Thus $P \in C_1$ by Proposition 1.3. 

Suppose, by induction, that $D^k(K[X])=C_k$ for $k \leq m-1$. 

Let $P \in D^m(K[X])$. 

Write $P_i=[P,x_i]$. 

Since $P$ has order $k \leq m-1$, it follows that $P_i \in C_{m-1}$. 

But, for all $1 \leq i,j \leq n$,
\begin{equation}
[P_i,x_j] = [[P,x_i],x_j] = [[P,x_j],x_i] = [P_j,x_i].
\end{equation}

Thus by Lemma 2.2 there exists $Q \in C_m$ such that $[Q,x_i]=P_i$, $1 \leq i \leq n$. 

Hence $[Q-P,x_i]=0$ in $D(K[X])$. 

Since this holds whenever $1 \leq i \leq n$, we conclude by Lemma 2.1 that $Q-P \in K[X]=C_0$. 

Therefore $P \in C_m$. 

Hence $D^m(K[X]) \subseteq C^m$, as we wanted to prove.

The ring of differential operators $D(R)$ of a commutative domain $R$ is not always generated by $R$ and $\text{Der}_K(R)$; see Exercise 3.8. 

However, this is true if the ring $R$ is regular; for example, if it is the coordinate ring of a non-singular irreducible affine variety. 

For a proof of this, see [McConnell and Robson 87, Ch.15]. 

The ring of Exercise 3.8 is the coordinate ring of a {\color{red}singular} curve, the cusp.


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% Section 3
\section{Exercises}
\begin{frame}[allowframebreaks]{C. }

\vspace{-0.5cm}

\textbf{3.1} Let $\mathcal{J}$ be a right ideal of a ring $R$. 

The {\color{red}idealizer} of $\mathcal{J}$ in $R$ is the set $\mathbb{I}(\mathcal{J})=\{a \in R : a\mathcal{J} \subseteq \mathcal{J}\}$. 

Show that $\mathbb{I}(\mathcal{J})$ is the largest subring of $R$ that contains $\mathcal{J}$ as a two-sided ideal.

\newpage

\textbf{3.2} Let $\mathcal{J}$ be an ideal of $S=K[x_1,\ldots,x_n]$. 
Let $\theta \in \text{End}_K(S)$.

\begin{enumerate}

\item Show that the formula $\overline{\theta}(\overline{f})=\overline{\theta(f)}$ defines a $K$-endomorphism of $S/\mathcal{J}$ if and only if $\theta(\mathcal{J}) \subseteq \mathcal{J}$. 
In that case, show that $\overline{\theta}=0$ if and only if $\theta(S) \subseteq \mathcal{J}$.

\item Show that if $\theta \in A_n \subseteq \text{End}_K(S)$ and $\theta(\mathcal{J}) \subseteq \mathcal{J}$, then $\overline{\theta} \in D(S/\mathcal{J})$, the ring of differential operators of the quotient $S/\mathcal{J}$. 
Show also that the order of $\overline{\theta}$ cannot exceed the order of $\theta$.

\item Show that if $\theta$ is a derivation of $S$ then $\overline{\theta}$ is a derivation of $S/\mathcal{J}$.
    
\end{enumerate}

\newpage

\textbf{3.3} Let $\mathcal{J}$ be an ideal of $S=K[x_1,\ldots,x_n]$. 

Let $P \in A_n$. 

Show that:
\begin{enumerate}
    \item If $P(S) \subseteq \mathcal{J}$ then $P \in \mathcal{J} A_n$.
    \item If $P(\mathcal{J}) \subseteq \mathcal{J}$ then $P \in \mathbb{I}(\mathcal{J} A_n)$.
\end{enumerate}

\newpage

\textbf{3.4} Let $\mathcal{J}$ be an ideal of $S=K[x_1,\ldots,x_n]$. 

Let $D(S/\mathcal{J})$ be the ring of differential operators of the quotient ring $S/\mathcal{J}$. 

Show that there is an injective ring homomorphism from $\mathbb{I}(\mathcal{J} A_n)/\mathcal{J} A_n$ into $D(S/\mathcal{J})$.

Hint: There is a homomorphism from $\mathbb{I}(\mathcal{J} A_n)$ to $D(S/\mathcal{J})$ which maps $\theta$ to $\overline{\theta}$, in the notation of Exercise 3.2. 

Now apply Exercise 3.3.


\newpage

\textbf{3.5} Let $\mathcal{J}$ be an ideal of $S=K[x_1,\ldots,x_n]$. 
%%
Let $\overline{f}$ be the image of $f \in S$ in the quotient ring $S/\mathcal{J}$. 

Suppose that $D$ is a derivation of $S/\mathcal{J}$ and choose $g_i \in S$ such that $\overline{g_i}=D(\overline{x_i})$, for $1 \leq i \leq n$.

(1) Show that if $B=\sum_{1}^{n} g_i \partial_i$, then $\overline{B}=D$, in the notation of Exercise 3.2.

(2) Let $\text{Der}_{\mathcal{J}} (S)$ be the set of derivations $D \in \text{Der}(S)$ such that $D(\mathcal{J}) \subseteq \mathcal{J}$. 
%%
Conclude, using Exercise 3.2, that there is an isomorphism of vector spaces between $\text{Der}_{\mathcal{J}} (S)/\mathcal{J}\text{Der}_{\mathcal{J}} (S)$ and $\text{Der}(S/\mathcal{J})$.

\newpage

\textbf{3.6} Let $R=K[t\pow{2},t\pow{3}]$.

(1) Show that $R$ is isomorphic to $K[x,y]/\mathcal{J}$, where $\mathcal{J}$ is the ideal of $K[x,y]$ generated by $y\pow{2}-x\pow{3}$.

(2) Let $D_1=2y\partial_x+3x\pow{2}\partial_y$ and $D_2=3y\partial_y-2x\partial_x$. 
%%
Use the previous exercise to show that the set of derivatives of $K[x,y]/\mathcal{J}$ is generated by $D_1$ and $D_2$ as a module over $K[x,y]/\mathcal{J}$.

(3) Conclude that $\text{Der}_K(R)$ is an $R$-module generated by $t\partial_t$ and $t\pow{2}\partial_t$, where $\partial_t=d/dt$.

\newpage

\textbf{3.7} Let $R$ be a commutative domain with field of fractions $Q$. 

Show that the set $\{P \in D(Q): P(R) \subseteq R\}$ is a subring of $D(R)$.

\newpage

\textbf{3.8} Let $R=K[t\pow{2},t\pow{3}]$. 

Let $\partial=d/dt$ and let $B_1$ be the $K$-algebra generated by $K(t)$ and $\partial$; see Exercise 2.4.3.

(1) Using Exercise 3.7, show that the following elements of $B_1$ belong to $D(R)$: $\partial\pow{2} - 2t\pow{-1}\partial$, $t\partial\pow{2} - \partial$ and $\partial\pow{3} - 3t\pow{-1}\partial\pow{2} + 3t\pow{-2}\partial$.

(2) Show that these elements do not belong to the subring of $B_1$ generated by $R$ and its derivations.

(3) Conclude that $D(R)$ is {\color{red}not} generated by $R$ and $\text{Der}_K(R)$.



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